mboxler Posted December 23, 2017 Share Posted December 23, 2017 I would like to measure the impedance of the K-33 in my K-horns at 400 hz. I've assembled a "patch cable" to place between my amp and the bass bin. The cable contains a 101.5 ohm 10 watt resistor that will end up in series with the woofer. The resistor actually measures 101.5 ohms. I have a free product on my Windows PC (fg_lite from Marchand Electronics) that allows me to generate the 400 hz signal. The chain will be PC-->DAC-->amplifier-->speaker cable-->patch cable-->right K-horn bass bin binding posts. These binding posts are connected directly to the K-33. When the signal is generated, I will turn the amp up until the tone becomes somewhat loud. I will use my Fluke 115 Trus RMS meter to measure the voltage across the right amplifier binding posts. I'll call the Vin. I will then measure the voltage drop across the 101.5 ohm resistor. I'll call this Vr. I will then calculate the current through the resistor. This will be Ir. Ir = Vr / 101.5. I will then calculate the voltage drop across the K-33, which should be Vin - Vr. I'll call that Vk33. Since the current passing through the resistor must be equal to the current through the K-33, I can calculate the impedance (resistance) of the K-33, I'll call that Rk33. Rk33 = Vk33 / Ir. This is what I ended up with... Vin = 6.325 volts Vr = 5.995 volts vk33 = .33 volts (man that's loud!) ir = .059 amps Vk33 = 5.887 ohms Does this method correctly measure the impedance of the K-33 at 400 hz? If so, I think I'll plot a few other frequencies just for fun. Sorry if this has been tried before. Thanks, Mike 1 Quote Link to comment Share on other sites More sharing options...
Deang Posted December 23, 2017 Share Posted December 23, 2017 I'm glad you started this thread, since it relates to something else that has come up again. I'm not a math person, it doesn't take much to overwhelm me. This is John Warren's analysis. http://www.northreadingeng.com/Klipschorn_model/basshorn_model.htm 1 Quote Link to comment Share on other sites More sharing options...
mboxler Posted December 23, 2017 Author Share Posted December 23, 2017 19 minutes ago, Deang said: I'm glad you started this thread, since it relates to something else that has come up again. I'm not a math person, it doesn't take much to overwhelm me. This is John Warren's analysis. http://www.northreadingeng.com/Klipschorn_model/basshorn_model.htm You caught on to my motivation. I saw that plot. Perhaps I'll try this at 50 hz and see if I'm still in the ballpark. Quote Link to comment Share on other sites More sharing options...
JohnA Posted December 23, 2017 Share Posted December 23, 2017 Fun experiment. I thought I had Al K's impedance plots of his Belle Klipsch, but I can't find them. I predict 6.3 ohms at 400 Hz. Quote Link to comment Share on other sites More sharing options...
Deang Posted December 23, 2017 Share Posted December 23, 2017 With a first order low pass in a horn loaded application, I just don't think it matters all that much. Al and me have always been in agreement about this. The horn is largely responsible for the acoustic crossover point. Quote Link to comment Share on other sites More sharing options...
Deang Posted December 23, 2017 Share Posted December 23, 2017 Need Warren in here. Chris A. and Gil would be cool too. More interesting and confusing stuff. https://www.klippel.de/fileadmin/klippel/Files/Know_How/Literature/Papers/Klippel_Nonlinearity_Poster.pdf I sometimes think the hardest part in all of this is trying to figure out what question to ask. Quote Link to comment Share on other sites More sharing options...
ClaudeJ1 Posted December 23, 2017 Share Posted December 23, 2017 10 minutes ago, Deang said: With a first order low pass in a horn loaded application, I just don't think it matters all that much. Al and me have always been in agreement about this. The horn is largely responsible for the acoustic crossover point. Coupled with the inductance, mass, etc. rolloff per Keele, yes indeed. But yes, the horn also has too many bends for the highs to escape. Quote Link to comment Share on other sites More sharing options...
mboxler Posted December 23, 2017 Author Share Posted December 23, 2017 I'm bummed. At 60 Hz, I get around 4 ohms. At 1000 hz I get around 7 ohms. Could my K-horn's not be sealed well enough against the wall, or did I just waste time with this technique. Quote Link to comment Share on other sites More sharing options...
JohnA Posted December 23, 2017 Share Posted December 23, 2017 Previous tests. I don't think a 1 ohm variation is significant. The Re of K-33-Es vary from 3.2 to 3.6 ohms, in my experience. Quote Link to comment Share on other sites More sharing options...
mboxler Posted December 23, 2017 Author Share Posted December 23, 2017 18 minutes ago, JohnA said: Previous tests. I don't think a 1 ohm variation is significant. The Re of K-33-Es vary from 3.2 to 3.6 ohms, in my experience. Thanks for doing that! Odd that the impedance is lower at 50 hz in the horn than in free air (which I assume is the black graph). Mike Quote Link to comment Share on other sites More sharing options...
Deang Posted December 23, 2017 Share Posted December 23, 2017 This is kind of cool. Only understand about half of it though. https://www.klippel.de/fileadmin/klippel/Files/Know_How/Literature/Papers/Klippel_Nonlinearity_Poster.pdf Quote Link to comment Share on other sites More sharing options...
WMcD Posted December 24, 2017 Share Posted December 24, 2017 For what its worth here is the impedance of a K-33 in a Belle. 1 Quote Link to comment Share on other sites More sharing options...
tigerwoodKhorns Posted December 24, 2017 Share Posted December 24, 2017 On 12/23/2017 at 10:35 AM, Deang said: I'm glad you started this thread, since it relates to something else that has come up again. I'm not a math person, it doesn't take much to overwhelm me. This is John Warren's analysis. http://www.northreadingeng.com/Klipschorn_model/basshorn_model.htm I'm glad that this is in here. I don't have the time to follow these threads closely and go over the math, but I am a nerd and miss this. My general question is how can a 2.4 mH inductor work in the belles La Scala and K Horn? My guess is that the cabinet changes the resistance of the woofer resulting in a higher crossover point in the smaller horns. I didn't check the math but it might be the opposite. I really miss this stuff! Quote Link to comment Share on other sites More sharing options...
mboxler Posted December 24, 2017 Author Share Posted December 24, 2017 A first order filter is doesn't do much within a couple hundred Hz either side of resonance. If everything was perfect (2.4 mH inductor, 6 ohm load, 400 hz), the voltage to the K-33 is down 3 dB. At 500 Hz, it's down 4 db. At 600, around 5 dB. If both the K-33 and K-55 were perfectly crossed (first order) at 400 hz, they would each be down 3 dB. The voltage across both drivers combine at 400 hz, resulting in a 0 dB summed voltage. It would be a 3 dB bump when summed, but the voltages across each driver is 90 degrees apart. As you can imagine, even if the K-33 were crossed at 400 hz and the K-55 at 500 hz, the combination would be down less than .5 dB around 450 hz (I didn't do the math). Too little to make any difference. Again, this is voltage only. Driver sensitivity, cabinet, room, and kids will change everything. That's my theory anyway! Mike Quote Link to comment Share on other sites More sharing options...
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